CODEWITHSUNDEEP
typesgo
cc young
cc young

Reputation: 20203

In go, how do you create an interface when methods are called by *Type?

Attempting to create an interface, but methods have *Type, not Type receivers

APOLOGIZE: was sleepy and mis-read error messages. Thought I was being block from creating the DB interface when in reality I was mis-using it. Sorry about that... will be more careful in the future!

type Char string

func (*Char) toType(v *string) interface{} {
        if v == nil {
                return (*Char)(nil)
        }
        var s string = *v
        ch := Char(s[0])
        return &ch
}
func (v *Char) toRaw() *string {
        if v == nil {
                return (*string)(nil)
        }
        s := *((*string)(v))
        return &s
}

from here I would like an interface that contains the methods toType and toRaw

type DB interface{
        toRaw() *string
        toType(*string) interface{}
}

does not work since the function receivers are pointers. I say this because when I try to use it I get the error.k

    Char does not implement DB (toRaw method requires pointer receiver)

Is there a way to create an interface from toType and toRaw, or do I need to backup and have the receivers be the types themselves and not pointers to types?

Upvotes: 6

Views: 2957

Answers (2)

newacct
newacct

Reputation: 122419

I don't understand what your problem is. Yes, the way you've written it, *Char conforms to the interface DB and Char doesn't. You can either

  1. change your code so that the methods operate on the non-pointer type Char directly (which will automatically also work for *Char too)
  2. only use *Char when you need something to be compatible with type DB

Upvotes: 3

Asgeir
Asgeir

Reputation: 1112

If you define your interface methods for the pointer type you must pass a pointer to the methods/functions expecting the interface.

Upvotes: 5

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